Question

A force $F=-K(yi+aj)$ (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0 $\therefore \overrightarrow{F}=-kx\hat{j}$

i.e. force is in negative y-direction while displacement is in positive x-direction.

$\therefore W_1=0$

Because force is perpendicular to displacement

Then particle moves from (a,0) to (a,a) along a line parallel to y-axis (x = +a) during this $\overrightarrow{F}=-k(y\hat{i}+a\hat{j})$

The first component of force, $-ky\hat{i}$ will not contribute any work because this components is along negative x-direction ($-\hat{i})$ while displacement is in positive y-direction (a,0) to (a,a).

The second component of force i.e. $-ka\hat{j}$

$\therefore W_2=(-ka\hat{j})\left(a\hat{j}\right)=(-ka)\left(a\right)=-k{a}^2$

So net work done on the particle $W=W_1+W_2$

$=0+(-k{a}^2)=-k{a}^2$